\newproblem{lay:1_1_18}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 1.1.18}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
  Do the three planes $2x_1+4x_2+4x_3=4$, $x_2-2x_3=-2$ and $2x_1+3x_2=0$ have at least one common point of intersection? Explain.
}{
   % Solution
	Let us construct the augmented system matrix
	\begin{center}
		$\left(\begin{array}{rrr|r}
		   2 &  4 &  4 &  4 \\
		   0 &  1 & -2 & -2 \\
		   2 &  3 &  0 &  0 \\
		\end{array}\right)$
	\end{center}
	Now, we apply row operations to determine whether it is compatible or not
	\begin{center}
		\begin{tabular}{cc}
			 $\mathbf{r}_3\leftarrow \mathbf{r}_3-\mathbf{r}_1$ &
				$\left(\begin{array}{rrr|r}
					 2 &  4 &  4 &  4 \\
					 0 &  1 & -2 & -2 \\
					 0 & -1 & -4 & -4 \\
				\end{array}\right)$ \\
			 $\mathbf{r}_3\leftarrow \mathbf{r}_3+\mathbf{r}_2$ &
				$\left(\begin{array}{rrr|r}
					 2 &  4 &  4 &  4 \\
					 0 &  1 & -2 & -2 \\
					 0 &  0 & -6 & -6 \\
				\end{array}\right)$ \\
		\end{tabular}
	\end{center}
  The system is compatible determinate and consequently the three planes intersect at a single point.
}
\useproblem{lay:1_1_18}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
